✔ 最佳答案
(設a不為0)
f_n(x)=f[f_n-1(x)], 故f_n(x)為有理式(分式)
設f_n(x)=P_n(x)/Q_n(x), 則
Q_0(x)=1, P_0(x)=x
Q_1(x)=1-x, P_1(x)=a
f_(n+1)(x)= a/{1- [P_n(x)/Q_n(x)]}= aQ_n(x)/[Q_n(x) - P_n(x)], 則
P_(n+1) (x)=a Q_n(x) ----(1)
Q_(n+1) (x)=Q_n(x) - P_n(x) ---(2)
則Q_(n+1) (x)= Q_n(x) - a Q_(n-1)(x)
Q_(n+1) (x)=Q_n(x) - a Q_(n-1) (x), Q_0(x)=1, Q_1(x)=1-x為2階遞迴
設α,β為二次方程式 m^2-m +a=0之二根
α=[1+√(1-4a)]/2, β=[1-√(1-4a)]/2, 則
Q_n(x)=Aα^n+Bβ^n
n=0: A+B=1
n=1: αA+βB=1-x,則
A=(1-x-β)/(α-β), B=(α-1+x)/(α-β)
Q_n(x)=[(1-x-β)α^n+(α-1+x)β^n]/(α-β)
P_n(x)=a Q_(n-1) (x)=a[(1-x-β)α^(n-1)+(α-1+x)β^(n-1)]/(α-β)
f_n(x)=P_n(x)/Q_n(x)
M.I.
n=1: f_1(x)=a/(1-x)=P_1(x)/Q_1(x) 成立
設f_n(x)=P_n(x)/Q_n(x), 則
f_(n+1) (x)= a/[1- f_n(x)]= a Q_n(x)/[Q_n(x) - P_n(x)] (by (1)&(2) )
= P_(n+1) (x)/ Q_(n+1) (x)
由數學歸納法,得證.